Find the characteristic equation of the given symmetric matrix,and then by inspection determine the dimensions of the eigenspaces.[6 3 31A = |3 6 3[3 3 6]The characteristic= 0equation of matrix A isLet A1 < A2.The dimension of the eigenspaceof A corresponding to A1 is equal toThe dimension of the eigenspaceof A corresponding to A, is equal to

Given matrix is A=633363336 First we will find the characteristic equation of the given matrix A=633363336 The characteristic equation of the matrix A is given by detA−λI=0 Now, detA−λI=0det633363336−λ100010001=0det6−λ3336−λ3336−λ=0−λ3+18λ2−81λ+108=0 Hence, characteristic equation of the matrix A is −λ3+18λ2−81λ+108=0Now, we will find the dimension of eigenspace of corresponding eigen values. Solve the equation −λ3+18λ2−81λ+108=0, we get λ=3,3,12 Let, λ1=3 and λ2=12 Given, λ1<λ2 Now, for λ1=3 solve A−λ1IX=0 633363336−3100010001x1x2x3=0333333333x1x2x3=0111000000x1x2x3=0x1+x2+x3=0 Let x2=t and x3=s Then x1=-t-s Hence, x1x2x3=−t−stsx1x2x3=−110t+−101s=−110,−101 Therefore, the dimension of eigenspace of matrix A corresponding eigen value λ1=3 is equal to 2.Now, for λ2=12 solve A−λ2IX=0 633363336−12100010001x1x2x3=0-6333-6333-6x1x2x3=010-101-1000x1x2x3=0 Hence, x1-x3=0....1x2-x3=0.....2 Let x2=t and x3=t Then x1=t Hence, x1x2x3=tttx1x2x3=111t=111 Therefore, the dimension of eigenspace of matrix A corresponding eigen value λ2=12 is equal to 1.